# Kirchhoff’s first rule – problems and solutions

1. Based on figure as shown below, if I_{1} = 5A, I_{2} = 3 A, I_{3} = 6 A, then I_{4 }= ………

Solution :

Kirchhoff’s first rule states that the sums of all currents entering the junction must equal the sum of all currents leaving the junction.

I_{1 }and I_{2 }are entering the junction, whereas I_{3} and I_{4 }are leaving.

Apply Kirchhoff’s first rule :

I_{1} + I_{2 }= I_{3} + I_{4}

5 + 3 = 6 + I_{4}

8 = 6 + I_{4}

I_{4} = 8 – 6

I_{4} = 2 Ampere

2. If I_{1} = I_{3} then I_{4} =……..

__Known :__

I_{1 }= I_{3} = 2 Ampere

I_{2} = 3 Ampere

I_{5 }= 4 Ampere

__Wanted:__ I_{4}

__Solution :__

I_{1 }and I_{4 }are entering the junction, whereas I_{2}, I_{3}, and I_{5 }are leaving.

Apply Kirchhoff’s first rule :

I_{1} + I_{4 }= I_{2} + I_{3 }+ I_{5}

2 + I_{4 }= 3 + 2 + 4

2 + I_{4 }= 9

I_{4 }= 9 – 2

I_{4} = 7 Ampere

3. I_{1} = …….

Solution :

I am entering the junction, whereas I_{1}, I_{2}, and I_{3} are leaving.

Apply Kirchhoff’s first rule :

I = I_{1} + I_{2 }+ I_{3}

7 A = I_{1} + 1 A + 4 A

7 A = I_{1} + 5 A

7 A – 5 A = I_{1}

I_{1 }= 2 Ampere

4. I_{2} = ………

__Wanted:__ I_{2}

__Solution :__

I_{1 }is entering the junction, whereas I_{2,} I_{3, }and I_{4} are leaving.

Apply Kirchhoff’s first rule :

I_{1} = I_{2} + I_{3} + I_{4}

6 = I_{2} + 3 + 1

6 = I_{2} + 4

I_{2 }= 6 – 4

I_{2} = 2 Ampere

5. If I_{2} = 1/4 I_{1} , then I_{3} = ………

__Known :__

I_{1} = 600 milliAmpere

__Wanted:__ I_{3}

__Solution :__

I_{2} = 1/4 I_{1 }= 1/4 (600 mA) = 150 mA

I_{1 }is entering the junction, whereas I_{2} and I_{3} are leaving.

Apply Kirchhoff’s first rule :

I_{1} = I_{2} + I_{3}

600 mA = 150 mA + I_{3}

I_{3} = 600 mA – 150 mA

I_{3} = 450 mA

6. Based on figure below, find I!

Solution :

Apply Kirchhoff‘s first rule :

3 + 5 + 2 = 7 + I

10 = 7 + I

10 – 7 = I

I = 3 Ampere

7. Based on figure below, if I_{1} = 4 A, I_{2} = 3 A, I_{4} = 7 A, I_{5} = 4 A, determine I_{3}.

Solution :

*Electric current entering the junction** = I*_{1 }*+ I*_{3 }*= 4 A + I*_{3}

*Electric current leaving the junction **= I*_{2 }*+ I*_{4 }*+ I*_{5}* = 3 A + 7 A + 4 A = 14 A*

14 A = 4 A + I_{3}

I_{3} = 14 A – 4 A

I_{3} = 10 A